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As in the title, I'm trying to find all the possible solutions to the equation: $x^2+x=0$ in $\mathbb Z_6$.

I know that the solution for the congruence equation is valid only if $m |(b-a)$. I assume that the equation is already in that form, so the solutions will be $\{0,2,3,5\}$. Is that correct? If not, I would appreciate any form of help.

Thank you.

user306691
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    There are only $6$ possible values for $x$, so you could test them all. Or, you could factorise $x^2 + x$ as $x(x + 1)$, and use the fact that a number is divisible by $6$ if and only if it is divisible by $2$ and divisible by $3$. You've given the right answer, but in too ambiguous a notation, because in theory, the sequence $2, 3, 5, \ldots$ could be continued in several different ways, even though it's almost 'obvious' what you mean. As I see someone has just noted in an answer, you haven't defined the symbols $m$, $a$, or $b$; you probably wanted to say something about divisibility by $6$. – Calum Gilhooley Mar 02 '16 at 21:19
  • What I was trying to say is that numbers $a$ and $b$ are congruent to each other, if and only if $a-b$ is divisible by $m$, where $m$ is the modulus. I'm not sure if this clarifies my question. – user306691 Mar 02 '16 at 21:22
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    "even though it's almost 'obvious' what you mean" It's not "almost' obvious'" obvious to me. I have no idea whatsoever what {2,3,5,...} is supposed to mean. Primes? And I don't know what "equation is already in that form" means. What form? 6 must divide x(x+1) and so either 6|x or 6|x+1 or 2|x and 3|x+1 or 3|x and 2|x+1. As we need only consider 0$\le$ x < 6 there are few solutions. Perhaps a better way of putting it is x(x+1) = 0 so x and x +1 are "zero divisors"; how many "zero divisors" are there in $\mathbb Z_6$? – fleablood Mar 02 '16 at 21:27
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    Yes, you should either insert some such clarification into the question, or else simply take the definition of congruence as read, because it's so well-known. It depends on how you finally choose to phrase your argument. (If you want to make an explicit deduction from the definition, you might want to state it explicitly.) By the way, I should have mentioned that your list of solutions may or may not be incomplete, depending on how exactly you mean the "$\ldots$" to continue! – Calum Gilhooley Mar 02 '16 at 21:29
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    @fleablood I meant that I could see what (s)he was getting at. But of course I agree that the meaning of the notation is not 'obvious' enough for it to serve to serve as a proper written solution. (The margin is too small to contain the wonderfully detailed critical comment I would like to have made.) :) – Calum Gilhooley Mar 02 '16 at 21:36
  • Can I say that the solution consists all the numbers of the form ${{2+km,3+km,5+km}}$, for any integer $k$?. Would that be more clear? Btw, I'm sorry for the confusion. – user306691 Mar 02 '16 at 21:39
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    I think you are worrying too much about presenting the answer as numbers. The question is about equivalency classes and there are only 6 of them. x = 0,1,2,3,4,5. That's all you have to express them as. So you can simply say the solutions are {2,3,5} (which is a subset of $\mathbb Z_6$) [ and "2" is understood to be defined as 2 = {$n \in \mathbb Z| n = 2 + k*6; k \in \mathbb Z$}]. But, by the way, you don't have them all. – fleablood Mar 02 '16 at 21:47
  • Forgot about the 0. Thanks for pointing out. I will try reformulate my answer based on what you all said. Thanks for helping me. – user306691 Mar 02 '16 at 21:52
  • Depending on what course you're following, and/or what book you're reading, you might want to introduce an explicit notation for elements of $\mathbb{Z}_6$, e.g. $[3] = { 3 + 6k : k \in \mathbb{Z} }$. Later, when you're more comfortable with congruence classes, and perhaps equivalence classes in general, you might more safely and comfortably use simply '$x$' to mean either an element of $\mathbb{Z}$ or an element of $\mathbb{Z}_6$, depending on context. Given the feeling of confusion you express (one which I often share!), it's probably safest at this stage to be completely explicit. – Calum Gilhooley Mar 02 '16 at 21:56
  • I'll definitely use the explicit notation. Thanks again. I really appreciate your help. – user306691 Mar 02 '16 at 22:05
  • It is confusing. When I learned I thought of Z6 as simply 6 elements with circular addition. It was so easy! But I missed out on all the power that it told may about entire sets of numbers. I think, in hindsight, I learned it the "wrong" way, but the math and arithmetic were fine. Once you learn it, it is the best of both worlds. – fleablood Mar 02 '16 at 22:10

2 Answers2

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Preliminary: $\mathbb Z_6$ = {0,1,2,3,4,5} where each $i$ is understood to represent an entire set of integers such that $i = \{n \in \mathbb Z| n = i + k*6; \text{ for some } k \in \mathbb Z\}$. That may be a little abstract and obtuse. In practical terms we can think of $\mathbb Z_6$ as the 6 digits with "circular" addition. But that's not really correct.

So $x^2 + x \equiv x(x+1) \equiv 0 \mod 6$. Thus $x$ and $x+1$ are the "zero divisors" of $\mathbb Z_6$ . $jk = 0 \mod 6$ means $6|jk$. Either $j$ or $k$ equal 0 (or 6) is an obvious solution but also as $6 = 2*3$, either $2|j$ and $3|k$ or $2|k $ and $3|j$ are ohter possibilities.

So we have either:

$x = 0$ or $x+1 = 0$. That is $x = 0$ or $x = 5$

or $2|x$ and $3|x+1$. That is $x =0,2,4$ and $x+1 = 0,3$ or in other words $x = 2$.

or $3|x$ and $2|x+1$. That is $x=0,3$ and $x+1=0,2,4$ or in other words $x =3$.

So the solutions are x = {0,2,3,5}. The only non-solutions are x $\ne$ {1,4}.

fleablood
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Maybe when you mention $b$ and $a$ you are thinking of something you learned about linear congruences? There are no $a$, $b$ mentioned in your post.

One salient observation is that you need only consider congruence classes; that is, $x=1$ will be a solution if and only if $7,\,13,$ etc. are. Thus, you only need to check finitely many integers (in fact, just six of them).

D. Wagner
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